Question

1. \sqrt3{dfrac{250f^{7}g^{3}}{2f^{2}g}}

Explanation:

Step1: Simplify the fraction inside the cube root

First, simplify the numerical part and the variables separately. For the numerical part, $\frac{250}{2} = 125$. For the variable \( f \), use the quotient rule of exponents \( \frac{a^m}{a^n}=a^{m - n} \), so \( \frac{f^7}{f^2}=f^{7 - 2}=f^5 \). For the variable \( g \), \( \frac{g^3}{g}=g^{3 - 1}=g^2 \). So now the expression inside the cube root is \( 125f^5g^2 \).
$$\sqrt[3]{\frac{250f^{7}g^{3}}{2f^{2}g}}=\sqrt[3]{125f^{5}g^{2}}$$

Step2: Factor the terms into perfect cubes and remaining factors

We know that \( 125 = 5^3 \), \( f^5=f^{3 + 2}=f^3\cdot f^2 \), and \( g^2 \) cannot be factored into a perfect cube (since the exponent is less than 3). So we can rewrite the expression inside the cube root as \( 5^3\cdot f^3\cdot f^2\cdot g^2 \).
$$\sqrt[3]{125f^{5}g^{2}}=\sqrt[3]{5^3\cdot f^3\cdot f^2\cdot g^2}$$

Step3: Apply the cube root property

The cube root of a product is the product of the cube roots, i.e., \( \sqrt[3]{abc}=\sqrt[3]{a}\cdot\sqrt[3]{b}\cdot\sqrt[3]{c} \). So we can take the cube root of the perfect cube terms: \( \sqrt[3]{5^3}=5 \), \( \sqrt[3]{f^3}=f \). The remaining terms \( f^2g^2 \) stay inside the cube root.
$$\sqrt[3]{5^3\cdot f^3\cdot f^2\cdot g^2}=5f\sqrt[3]{f^2g^2}$$

Answer:

\( 5f\sqrt[3]{f^{2}g^{2}} \)